Question

practice: 7. answer the following questions about this reaction: 2na(s) + 2h₂o(l) → 2naoh(aq) + h₂(g) a. identify the limiting reactant if you begin with 0.345 mol na and 0.678 mol h₂o. b. how much of the excess reactant, in grams, will be leftover? c. how much of each product, in grams, will be formed?

Explanation:

Step1: Determine mole - ratio from balanced equation

The balanced equation is $2Na(s)+2H_2O(l)
ightarrow2NaOH(aq) + H_2(g)$. The mole - ratio of $Na$ to $H_2O$ is $1:1$.

Step2: Calculate moles of $H_2O$ needed for $Na$ to react completely

If we have $n_{Na}=0.345$ mol of $Na$, the moles of $H_2O$ needed for complete reaction of $Na$ is $n_{H_2O\ needed}=0.345$ mol (since mole - ratio is $1:1$). We have $n_{H_2O\ available}=0.678$ mol. So, $Na$ is the limiting reactant.

Step3: Calculate moles of excess reactant left

Moles of $H_2O$ left, $n_{H_2O\ left}=n_{H_2O\ available}-n_{H_2O\ reacted}$. Since $n_{H_2O\ reacted}=0.345$ mol, $n_{H_2O\ left}=0.678 - 0.345=0.333$ mol. Mass of $H_2O$ left, $m_{H_2O}=n_{H_2O\ left}\times M_{H_2O}$, where $M_{H_2O}=18.015$ g/mol. So, $m_{H_2O}=0.333\times18.015 = 5.999$ g $\approx6.00$ g.

Step4: Calculate moles of products formed

From the balanced equation, mole - ratio of $Na$ to $NaOH$ is $1:1$ and mole - ratio of $Na$ to $H_2$ is $2:1$. Moles of $NaOH$ formed, $n_{NaOH}=0.345$ mol. Mass of $NaOH$, $m_{NaOH}=n_{NaOH}\times M_{NaOH}$, where $M_{NaOH}=40.0$ g/mol. So, $m_{NaOH}=0.345\times40.0 = 13.8$ g. Moles of $H_2$ formed, $n_{H_2}=\frac{0.345}{2}=0.1725$ mol. Mass of $H_2$, $m_{H_2}=n_{H_2}\times M_{H_2}$, where $M_{H_2}=2.016$ g/mol. So, $m_{H_2}=0.1725\times2.016 = 0.3488$ g $\approx0.349$ g.

Answer:

a. The limiting reactant is $Na$.
b. The mass of the excess reactant ($H_2O$) left over is approximately $6.00$ g.
c. The mass of $NaOH$ formed is $13.8$ g and the mass of $H_2$ formed is approximately $0.349$ g.